Answer
$$ y'= (4x-8)e^{(x-1)^2}e^{(x-3)^2}.$$
Work Step by Step
Taking the $\ln $ on both sides of the equation, we get
$$\ln y= \ln \left(e^{(x-1)^2}e^{(x-3)^2}\right)$$
Then using the properties of $\ln $, we can write
$$\ln y= \ln e^{(x-1)^2}+\ln e^{(x-3)^2}\\
=(x-1)^2+(x-3)^2.$$
Now taking the derivative for the above equation, we have
$$\frac{y'}{y}= 2(x-1)+2(x-3)=4x-8,$$
Hence $ y'$ is given by
$$ y'=y(4x-8)=(4x-8)e^{(x-1)^2}e^{(x-3)^2}$$
